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Chapter 3
Composition of Substances and
Solutions
Figure 3.1 The water in a swimming pool is a complex mixture of substances whose relative amounts must be carefully maintained to ensure the health and comfort of people using the pool. (credit: modification of work by Vic Brincat)
Chapter Outline
3.1 Formula Mass and the Mole Concept 3.2 Determining Empirical and Molecular Formulas 3.3 Molarity 3.4 Other Units for Solution Concentrations
Introduction
Swimming pools have long been a popular means of recreation, exercise, and physical therapy. Since it is impractical to refill large pools with fresh water on a frequent basis, pool water is regularly treated with chemicals to prevent the growth of harmful bacteria and algae. Proper pool maintenance requires regular additions of various chemical
compounds in carefully measured amounts. For example, the relative amount of calcium ion, Ca2+, in the water should be maintained within certain limits to prevent eye irritation and avoid damage to the pool bed and plumbing. To maintain proper calcium levels, calcium cations are added to the water in the form of an ionic compound that also
contains anions; thus, it is necessary to know both the relative amount of Ca2+^ in the compound and the volume of water in the pool in order to achieve the proper calcium level. Quantitative aspects of the composition of substances (such as the calcium-containing compound) and mixtures (such as the pool water) are the subject of this chapter.
3.1 Formula Mass and the Mole Concept
By the end of this section, you will be able to:
- Calculate formula masses for covalent and ionic compounds
- Define the amount unit mole and the related quantity Avogadro’s number Explain the relation between mass, moles, and numbers of atoms or molecules, and perform calculations deriving these quantities from one another
We can argue that modern chemical science began when scientists started exploring the quantitative as well as the qualitative aspects of chemistry. For example, Dalton’s atomic theory was an attempt to explain the results of measurements that allowed him to calculate the relative masses of elements combined in various compounds. Understanding the relationship between the masses of atoms and the chemical formulas of compounds allows us to quantitatively describe the composition of substances.
Formula Mass
In an earlier chapter, we described the development of the atomic mass unit, the concept of average atomic masses, and the use of chemical formulas to represent the elemental makeup of substances. These ideas can be extended to calculate the formula mass of a substance by summing the average atomic masses of all the atoms represented in the substance’s formula.
Formula Mass for Covalent Substances
For covalent substances, the formula represents the numbers and types of atoms composing a single molecule of the substance; therefore, the formula mass may be correctly referred to as a molecular mass. Consider chloroform (CHCl 3 ), a covalent compound once used as a surgical anesthetic and now primarily used in the production of the “anti-stick” polymer, Teflon. The molecular formula of chloroform indicates that a single molecule contains one carbon atom, one hydrogen atom, and three chlorine atoms. The average molecular mass of a chloroform molecule is therefore equal to the sum of the average atomic masses of these atoms. Figure 3.2 outlines the calculations used to derive the molecular mass of chloroform, which is 119.37 amu.
Figure 3.2 The average mass of a chloroform molecule, CHCl 3 , is 119.37 amu, which is the sum of the average atomic masses of each of its constituent atoms. The model shows the molecular structure of chloroform.
Likewise, the molecular mass of an aspirin molecule, C 9 H 8 O 4 , is the sum of the atomic masses of nine carbon atoms, eight hydrogen atoms, and four oxygen atoms, which amounts to 180.15 amu ( Figure 3.3 ).
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Figure 3.4 Table salt, NaCl, contains an array of sodium and chloride ions combined in a 1:1 ratio. Its formula mass is 58.44 amu.
Note that the average masses of neutral sodium and chlorine atoms were used in this computation, rather than the masses for sodium cations and chlorine anions. This approach is perfectly acceptable when computing the formula mass of an ionic compound. Even though a sodium cation has a slightly smaller mass than a sodium atom (since it is missing an electron), this difference will be offset by the fact that a chloride anion is slightly more massive than a chloride atom (due to the extra electron). Moreover, the mass of an electron is negligibly small with respect to the mass of a typical atom. Even when calculating the mass of an isolated ion, the missing or additional electrons can generally be ignored, since their contribution to the overall mass is negligible, reflected only in the nonsignificant digits that will be lost when the computed mass is properly rounded. The few exceptions to this guideline are very light ions derived from elements with precisely known atomic masses.
Example 3.
Computing Formula Mass for an Ionic Compound Aluminum sulfate, Al 2 (SO 4 ) 3 , is an ionic compound that is used in the manufacture of paper and in various water purification processes. What is the formula mass (amu) of this compound? Solution
The formula for this compound indicates it contains Al3+^ and SO 4 2−^ ions combined in a 2:3 ratio. For purposes of computing a formula mass, it is helpful to rewrite the formula in the simpler format, Al 2 S 3 O 12. Following the approach outlined above, the formula mass for this compound is calculated as follows:
Check Your Learning Calcium phosphate, Ca 3 (PO 4 ) 2 , is an ionic compound and a common anti-caking agent added to food products. What is the formula mass (amu) of calcium phosphate? Answer: 310.18 amu
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The Mole
The identity of a substance is defined not only by the types of atoms or ions it contains, but by the quantity of each type of atom or ion. For example, water, H 2 O, and hydrogen peroxide, H 2 O 2 , are alike in that their
respective molecules are composed of hydrogen and oxygen atoms. However, because a hydrogen peroxide molecule contains two oxygen atoms, as opposed to the water molecule, which has only one, the two substances exhibit very different properties. Today, we possess sophisticated instruments that allow the direct measurement of these defining microscopic traits; however, the same traits were originally derived from the measurement of macroscopic properties (the masses and volumes of bulk quantities of matter) using relatively simple tools (balances and volumetric glassware). This experimental approach required the introduction of a new unit for amount of substances, the mole , which remains indispensable in modern chemical science.
The mole is an amount unit similar to familiar units like pair, dozen, gross, etc. It provides a specific measure of the number of atoms or molecules in a bulk sample of matter. A mole is defined as the amount of substance containing the same number of discrete entities (such as atoms, molecules, and ions) as the number of atoms in a sample of pure (^12) C weighing exactly 12 g. One Latin connotation for the word “mole” is “large mass” or “bulk,” which is consistent
with its use as the name for this unit. The mole provides a link between an easily measured macroscopic property, bulk mass, and an extremely important fundamental property, number of atoms, molecules, and so forth.
The number of entities composing a mole has been experimentally determined to be 6.02214179 × 1023 , a
fundamental constant named Avogadro’s number ( NA ) or the Avogadro constant in honor of Italian scientist Amedeo
Avogadro. This constant is properly reported with an explicit unit of “per mole,” a conveniently rounded version
being 6.022 × 1023 /mol.
Consistent with its definition as an amount unit, 1 mole of any element contains the same number of atoms as 1 mole of any other element. The masses of 1 mole of different elements, however, are different, since the masses of the individual atoms are drastically different. The molar mass of an element (or compound) is the mass in grams of 1 mole of that substance, a property expressed in units of grams per mole (g/mol) (see Figure 3.5 ).
Figure 3.5 Each sample contains 6.022 × 1023 atoms —1.00 mol of atoms. From left to right (top row): 65.4 g zinc, 12.0 g carbon, 24.3 g magnesium, and 63.5 g copper. From left to right (bottom row): 32.1 g sulfur, 28.1 g silicon, 207 g lead, and 118.7 g tin. (credit: modification of work by Mark Ott)
Figure 3.7 The number of molecules in a single droplet of water is roughly 100 billion times greater than the number of people on earth. (credit: “tanakawho”/Wikimedia commons)
The mole is used in chemistry to represent 6.022 × 1023 of something, but it can be difficult to conceptualize such a large number. Watch this video (http://openstaxcollege.org/l/16molevideo) and then complete the “Think” questions that follow. Explore more about the mole by reviewing the information under “Dig Deeper.”
The relationships between formula mass, the mole, and Avogadro’s number can be applied to compute various quantities that describe the composition of substances and compounds. For example, if we know the mass and chemical composition of a substance, we can determine the number of moles and calculate number of atoms or molecules in the sample. Likewise, if we know the number of moles of a substance, we can derive the number of atoms or molecules and calculate the substance’s mass.
Example 3.
Deriving Moles from Grams for an Element According to nutritional guidelines from the US Department of Agriculture, the estimated average requirement for dietary potassium is 4.7 g. What is the estimated average requirement of potassium in moles? Solution The mass of K is provided, and the corresponding amount of K in moles is requested. Referring to the periodic table, the atomic mass of K is 39.10 amu, and so its molar mass is 39.10 g/mol. The given mass of
Link to Learning
K (4.7 g) is a bit more than one-tenth the molar mass (39.10 g), so a reasonable “ballpark” estimate of the number of moles would be slightly greater than 0.1 mol. The molar amount of a substance may be calculated by dividing its mass (g) by its molar mass (g/mol):
The factor-label method supports this mathematical approach since the unit “g” cancels and the answer has units of “mol:”
4.7 g K ⎛⎝ 39.10 gmol K^ ⎞⎠ = 0.12 mol K
The calculated magnitude (0.12 mol K) is consistent with our ballpark expectation, since it is a bit greater than 0.1 mol. Check Your Learning Beryllium is a light metal used to fabricate transparent X-ray windows for medical imaging instruments. How many moles of Be are in a thin-foil window weighing 3.24 g? Answer: 0.360 mol
Example 3.
Deriving Grams from Moles for an Element
A liter of air contains 9.2 × 10 −4^ mol argon. What is the mass of Ar in a liter of air?
Solution The molar amount of Ar is provided and must be used to derive the corresponding mass in grams. Since the amount of Ar is less than 1 mole, the mass will be less than the mass of 1 mole of Ar, approximately 40 g. The molar amount in question is approximately one-one thousandth (~10−3) of a mole, and so the corresponding mass should be roughly one-one thousandth of the molar mass (~0.04 g):
In this case, logic dictates (and the factor-label method supports) multiplying the provided amount (mol) by the molar mass (g/mol):
9.2 × 10−4^ mol Ar ⎛⎝39.95 g
mol Ar
⎠ = 0.037 g Ar
The result is in agreement with our expectations, around 0.04 g Ar. Check Your Learning What is the mass of 2.561 mol of gold? Answer: 504.4 g
Example 3.
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Deriving Moles from Grams for a Compound Our bodies synthesize protein from amino acids. One of these amino acids is glycine, which has the molecular formula C 2 H 5 O 2 N. How many moles of glycine molecules are contained in 28.35 g of glycine? Solution We can derive the number of moles of a compound from its mass following the same procedure we used for an element in Example 3.3 :
The molar mass of glycine is required for this calculation, and it is computed in the same fashion as its molecular mass. One mole of glycine, C 2 H 5 O 2 N, contains 2 moles of carbon, 5 moles of hydrogen, 2 moles of oxygen, and 1 mole of nitrogen:
The provided mass of glycine (~28 g) is a bit more than one-third the molar mass (~75 g/mol), so we would expect the computed result to be a bit greater than one-third of a mole (~0.33 mol). Dividing the compound’s mass by its molar mass yields:
28.35 g glycine ⎛⎝mol glycine75.07 g^ ⎞⎠ = 0.378 mol glycine
This result is consistent with our rough estimate. Check Your Learning How many moles of sucrose, C 12 H 22 O 11 , are in a 25-g sample of sucrose? Answer: 0.073 mol
Example 3.
Deriving Grams from Moles for a Compound Vitamin C is a covalent compound with the molecular formula C 6 H 8 O 6. The recommended daily dietary allowance of vitamin C for children aged 4–8 years is 1.42 × 10 −4^ mol. What is the mass of this allowance in grams? Solution As for elements, the mass of a compound can be derived from its molar amount as shown:
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The molar mass for this compound is computed to be 176.124 g/mol. The given number of moles is a very small fraction of a mole (~10−4^ or one-ten thousandth); therefore, we would expect the corresponding mass to be about one-ten thousandth of the molar mass (~0.02 g). Performing the calculation, we get:
1.42 × 10−4^ mol vitamin C ⎛⎝ 176.124 g
mol vitamin C
⎠ = 0.0250 g vitamin C
This is consistent with the anticipated result. Check Your Learning What is the mass of 0.443 mol of hydrazine, N 2 H 4? Answer: 14.2 g
Example 3.
Deriving the Number of Atoms and Molecules from the Mass of a Compound A packet of an artificial sweetener contains 40.0 mg of saccharin (C 7 H 5 NO 3 S), which has the structural formula:
Given that saccharin has a molar mass of 183.18 g/mol, how many saccharin molecules are in a 40.0-mg (0.0400-g) sample of saccharin? How many carbon atoms are in the same sample? Solution The number of molecules in a given mass of compound is computed by first deriving the number of moles, as demonstrated in Example 3.6 , and then multiplying by Avogadro’s number:
Using the provided mass and molar mass for saccharin yields:
0.0400 g C 7 H 5 NO 3 S ⎛⎝183.18 g C^ mol C^7 H^5 NO^3 S 7 H 5 NO 3 S
⎞ ⎠
⎛ ⎝
⎜6.022 × 10
23 C
7 H 5 NO 3 S molecules 1 mol C 7 H 5 NO 3 S
⎞ ⎠
⎟
= 1.31 × 10^20 C 7 H 5 NO 3 S molecules
One neurotransmitter that has been very extensively studied is dopamine, C 8 H 11 NO 2. Dopamine is involved in various neurological processes that impact a wide variety of human behaviors. Dysfunctions in the dopamine systems of the brain underlie serious neurological diseases such as Parkinson’s and schizophrenia.
Figure 3.10 (a) Chemical signals are transmitted from neurons to other cells by the release of neurotransmitter molecules into the small gaps (synapses) between the cells. (b) Dopamine, C 8 H 11 NO 2 , is a neurotransmitter involved in a number of neurological processes.
One important aspect of the complex processes related to dopamine signaling is the number of neurotransmitter molecules released during exocytosis. Since this number is a central factor in determining neurological response (and subsequent human thought and action), it is important to know how this number changes with certain controlled stimulations, such as the administration of drugs. It is also important to understand the mechanism responsible for any changes in the number of neurotransmitter molecules released—for example, some dysfunction in exocytosis, a change in the number of vesicles in the neuron, or a change in the number of neurotransmitter molecules in each vesicle. Significant progress has been made recently in directly measuring the number of dopamine molecules stored in individual vesicles and the amount actually released when the vesicle undergoes exocytosis. Using miniaturized probes that can selectively detect dopamine molecules in very small amounts, scientists have determined that the vesicles of a certain type of mouse brain neuron contain an average of 30,000 dopamine molecules per vesicle (about 5 × 10−20^ mol or 50 zmol). Analysis of these neurons from mice subjected to various drug therapies shows significant changes in the average number of dopamine molecules contained in individual vesicles, increasing or decreasing by up to three-fold, depending on the specific drug used. These studies also indicate that not all of the dopamine in a given vesicle is released during exocytosis, suggesting that it may be possible to regulate the fraction released using pharmaceutical therapies.[1]
- Omiatek, Donna M., Amanda J. Bressler, Ann-Sofie Cans, Anne M. Andrews, Michael L. Heien, and Andrew G. Ewing. “The Real Catecholamine Content of Secretory Vesicles in the CNS Revealed by Electrochemical Cytometry.” Scientific Report 3 (2013): 1447, accessed January 14, 2015, doi:10.1038/srep01447.
3.2 Determining Empirical and Molecular Formulas
By the end of this section, you will be able to:
- Compute the percent composition of a compound
- Determine the empirical formula of a compound
- Determine the molecular formula of a compound
In the previous section, we discussed the relationship between the bulk mass of a substance and the number of atoms or molecules it contains (moles). Given the chemical formula of the substance, we were able to determine the amount of the substance (moles) from its mass, and vice versa. But what if the chemical formula of a substance is unknown? In this section, we will explore how to apply these very same principles in order to derive the chemical formulas of unknown substances from experimental mass measurements.
Percent Composition
The elemental makeup of a compound defines its chemical identity, and chemical formulas are the most succinct way of representing this elemental makeup. When a compound’s formula is unknown, measuring the mass of each of its constituent elements is often the first step in the process of determining the formula experimentally. The results of these measurements permit the calculation of the compound’s percent composition , defined as the percentage by mass of each element in the compound. For example, consider a gaseous compound composed solely of carbon and hydrogen. The percent composition of this compound could be represented as follows:
% H = (^) mass compound mass H × 100 %
% C = (^) mass compound mass C × 100 %
If analysis of a 10.0-g sample of this gas showed it to contain 2.5 g H and 7.5 g C, the percent composition would be calculated to be 25% H and 75% C:
% H = (^) 10.0 g compound 2.5 g H × 100 % = 25 %
% C = (^) 10.0 g compound 7.5 g C × 100 % = 75 %
Example 3.
Calculation of Percent Composition Analysis of a 12.04-g sample of a liquid compound composed of carbon, hydrogen, and nitrogen showed it to contain 7.34 g C, 1.85 g H, and 2.85 g N. What is the percent composition of this compound? Solution To calculate percent composition, we divide the experimentally derived mass of each element by the overall mass of the compound, and then convert to a percentage:
% C = (^) 12.04 g compound 7.34 g C × 100 % = 61.0 %
% H = (^) 12.04 g compound 1.85 g H × 100 % = 15.4 %
% N = (^) 12.04 g compound 2.85 g N × 100 % = 23.7 %
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To three significant digits, what is the mass percentage of iron in the compound Fe 2 O 3? Answer: 69.9% Fe
Determination of Empirical Formulas
As previously mentioned, the most common approach to determining a compound’s chemical formula is to first measure the masses of its constituent elements. However, we must keep in mind that chemical formulas represent the relative numbers , not masses, of atoms in the substance. Therefore, any experimentally derived data involving mass must be used to derive the corresponding numbers of atoms in the compound. To accomplish this, we can use molar masses to convert the mass of each element to a number of moles. We then consider the moles of each element relative to each other, converting these numbers into a whole-number ratio that can be used to derive the empirical formula of the substance. Consider a sample of compound determined to contain 1.71 g C and 0.287 g H. The corresponding numbers of atoms (in moles) are:
1.17 g C × (^) 12.01 g C 1 mol C = 0.142 mol C
0.287 g H × (^) 1.008 g H 1 mol H = 0.284 mol H
Thus, we can accurately represent this compound with the formula C0.142H0.248. Of course, per accepted convention,
formulas contain whole-number subscripts, which can be achieved by dividing each subscript by the smaller subscript:
C (^) 0.
H 0.
or CH 2
(Recall that subscripts of “1” are not written but rather assumed if no other number is present.)
The empirical formula for this compound is thus CH 2. This may or not be the compound’s molecular formula as well;
however, we would need additional information to make that determination (as discussed later in this section).
Consider as another example a sample of compound determined to contain 5.31 g Cl and 8.40 g O. Following the same approach yields a tentative empirical formula of:
C10.150 O0.525 = Cl (^) 0.
O 0.
= ClO3.
In this case, dividing by the smallest subscript still leaves us with a decimal subscript in the empirical formula. To convert this into a whole number, we must multiply each of the subscripts by two, retaining the same atom ratio and yielding Cl 2 O 7 as the final empirical formula.
In summary, empirical formulas are derived from experimentally measured element masses by:
- Deriving the number of moles of each element from its mass
- Dividing each element’s molar amount by the smallest molar amount to yield subscripts for a tentative empirical formula
- Multiplying all coefficients by an integer, if necessary, to ensure that the smallest whole-number ratio of subscripts is obtained
Figure 3.11 outlines this procedure in flow chart fashion for a substance containing elements A and X.
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Figure 3.11 The empirical formula of a compound can be derived from the masses of all elements in the sample.
Example 3.
Determining a Compound’s Empirical Formula from the Masses of Its Elements A sample of the black mineral hematite ( Figure 3.12 ), an oxide of iron found in many iron ores, contains 34.97 g of iron and 15.03 g of oxygen. What is the empirical formula of hematite?
Figure 3.12 Hematite is an iron oxide that is used in jewelry. (credit: Mauro Cateb)
Solution For this problem, we are given the mass in grams of each element. Begin by finding the moles of each:
34.97 g Fe⎛⎝mol Fe55.85 g^ ⎞⎠ = 0.6261 mol Fe
15.03 g O⎛⎝ 16.00 gmol O^ ⎞⎠ = 0.9394 mol O
Next, derive the iron-to-oxygen molar ratio by dividing by the lesser number of moles:
0.6261 = 1.000 mol Fe
0.6261 = 1.500 mol O
Figure 3.13 An oxide of carbon is removed from these fermentation tanks through the large copper pipes at the top. (credit: “Dual Freq”/Wikimedia Commons)
Solution
Since the scale for percentages is 100, it is most convenient to calculate the mass of elements present in a sample weighing 100 g. The calculation is “most convenient” because, per the definition for percent composition, the mass of a given element in grams is numerically equivalent to the element’s mass percentage. This numerical equivalence results from the definition of the “percentage” unit, whose name is derived from the Latin phrase per centum meaning “by the hundred.” Considering this definition, the mass percentages provided may be more conveniently expressed as fractions:
27.29 % C = (^) 100 g compound27.29 g C
72.71 % O = (^) 100 g compound72.71 g O
The molar amounts of carbon and hydrogen in a 100-g sample are calculated by dividing each element’s mass by its molar mass:
27.29 g C⎛⎝ 12.01 gmol C^ ⎞⎠ = 2.272 mol C
72.71 g O⎛⎝ 16.00 gmol O^ ⎞⎠ = 4.544 mol O
Coefficients for the tentative empirical formula are derived by dividing each molar amount by the lesser of the two:
2.272 mol C 2.272 = 1 4.544 mol O 2.272 = 2
Since the resulting ratio is one carbon to two oxygen atoms, the empirical formula is CO 2.
Check Your Learning What is the empirical formula of a compound containing 40.0% C, 6.71% H, and 53.28% O? Answer: CH 2 O
Derivation of Molecular Formulas
Recall that empirical formulas are symbols representing the relative numbers of a compound’s elements. Determining the absolute numbers of atoms that compose a single molecule of a covalent compound requires knowledge of both its empirical formula and its molecular mass or molar mass. These quantities may be determined experimentally by various measurement techniques. Molecular mass, for example, is often derived from the mass spectrum of the compound (see discussion of this technique in the previous chapter on atoms and molecules). Molar mass can be measured by a number of experimental methods, many of which will be introduced in later chapters of this text.
Molecular formulas are derived by comparing the compound’s molecular or molar mass to its empirical formula mass. As the name suggests, an empirical formula mass is the sum of the average atomic masses of all the atoms represented in an empirical formula. If we know the molecular (or molar) mass of the substance, we can divide this by the empirical formula mass in order to identify the number of empirical formula units per molecule, which we designate as n :
molecular or molar mass ⎛⎝amu or (^) molg^ ⎞⎠ empirical formula mass ⎛⎝amu or (^) molg^ ⎞⎠
= n formula units/molecule
The molecular formula is then obtained by multiplying each subscript in the empirical formula by n , as shown by the generic empirical formula AxBy:
(Ax By)n = Anx Bnx
For example, consider a covalent compound whose empirical formula is determined to be CH 2 O. The empirical
formula mass for this compound is approximately 30 amu (the sum of 12 amu for one C atom, 2 amu for two H atoms, and 16 amu for one O atom). If the compound’s molecular mass is determined to be 180 amu, this indicates that molecules of this compound contain six times the number of atoms represented in the empirical formula:
180 amu/molecule (^30) formula unitamu
= 6 formula units/molecule
Molecules of this compound are then represented by molecular formulas whose subscripts are six times greater than those in the empirical formula:
(CH 2 O) 6 = C 6 H 12 O 6
Note that this same approach may be used when the molar mass (g/mol) instead of the molecular mass (amu) is used. In this case, we are merely considering one mole of empirical formula units and molecules, as opposed to single units and molecules.
Example 3.
Determination of the Molecular Formula for Nicotine Nicotine, an alkaloid in the nightshade family of plants that is mainly responsible for the addictive nature of cigarettes, contains 74.02% C, 8.710% H, and 17.27% N. If 40.57 g of nicotine contains 0.2500 mol nicotine, what is the molecular formula?
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